An electric range burner weighing 683.0 grams is turned off after reaching a temperature of 477.6°c, and is allowed to cool down to 23.2°c.

calculate the specific heat of the burner if all the heat evolved from the burner is used to heat 552.0 grams of water from 23.2°c to 80.3°c.

0.102 cal/g.°C

Explanation:

According to the law of conservation of energy, the sum of the heat released by the electric burner (Qb) and the heat absorbed by the water (Qw) is zero.

Qb + Qw = 0

Qb = -Qw

Both heats can be calculated using the following expression.

Q = c × m × ΔT

where,

c: specific heat

m: mass

ΔT: change in the temperature

Then,

Qb = -Qw

cb × mb × ΔTb = - cw × mw × ΔTw

cb = - cw × mw × ΔTw / mb × ΔTb

cb = - (1 cal/g.°C) × 552.0 g × (80.3°C - 23.2°C) / 683.0 g × (23.2°C - 477.6°C)

cb = 0.102 cal/g.°C

can you list the answers so i can answer

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.